First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:
2as = v² - u²
Because the final velocity v is 0 in such cases
s = -u²/2a; because both u and a are downwards, the negative sign cancels
s = 14.5² / 2*9.81
s = 10.72 meters
Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m
We will use the formula
s = ut + 0.5at²
to find the time taken with the initial velocity u = 0.
55.72 = 0.5 * 9.81 * t²
t = 3.37 seconds
I think the correct answer is C
Answer:
At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop
Explanation:
The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s
To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g
we write out the equation of motion thus.
S = ut + 0.5at²
wgere
S = distance to come to complete stop
u = final velocoty = 0 m/s
a = acceleration = 60g = 60 × 9.81
t = time = 36 ms
as can be seen, the above equation calls up the given variable as a function of the required variable thus
S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m
At 60g, a person will travel a distance of 0.38 m before coing to a complete stop
